The lower case sigma character (σ) represents standard deviation.
There is no such thing. Maybe your professor meant , Standard Deviation, The Mean. (2 different things.)
There is no such thing. The standard error can be calculated for a sample of any size greater than 1.
There is absolutely no relationship to what you've asked. I'm pretty sure you simply framed the question in the wrong way, but to literally answer your question... none. Zero relationship. There's no such thing. There is however a relationship between standard deviation and a CI, but a CI can in no shape way or form influence a standard deviation.
There doesn't exist such a thing. What does exist are standardized variables, which are variables with mean = 0 and standard deviation = 1
Standard error of the mean (SEM) and standard deviation of the mean is the same thing. However, standard deviation is not the same as the SEM. To obtain SEM from the standard deviation, divide the standard deviation by the square root of the sample size.
The lower case sigma character (σ) represents standard deviation.
No. But they are related. If a sample of size n is taken, a standard deviation can be calculated. This is usually denoted as "s" however some textbooks will use the symbol, sigma. The standard deviation of a sample is usually used to estimate the standard deviation of the population. In this case, we use n-1 in the denomimator of the equation. The variance of the sample is the square of the sample's standard deviation. In many textbooks it is denoted as s2. In denoting the standard deviation and variance of populations, the symbols sigma and sigma2 should be used. One last note. We use standard deviations in describing uncertainty as it's easier to understand. If our measurements are in days, then the standard deviation will also be in days. The variance will be in units of days2.
There is no such thing. Maybe your professor meant , Standard Deviation, The Mean. (2 different things.)
There is no such thing. The standard error can be calculated for a sample of any size greater than 1.
There is absolutely no relationship to what you've asked. I'm pretty sure you simply framed the question in the wrong way, but to literally answer your question... none. Zero relationship. There's no such thing. There is however a relationship between standard deviation and a CI, but a CI can in no shape way or form influence a standard deviation.
"Variance" and "Standard deviation" are numbers that describe a set of data that typically contains several numbers. Applied to a single number, neither of them has any meaning. -- The variance, standard deviation, and mean squared error of 7 are all zero. -- The mean, median, mode, average, max, min, RMS, and absolute value of 7 are all 7 . None of these facts tells you a thing about ' 7 ' that you didn't already know as soon as you found out that it was ' 7 '.
There doesn't exist such a thing. What does exist are standardized variables, which are variables with mean = 0 and standard deviation = 1
Standard deviation is the spread of the data. If each score has 7 added, this would not affect the spread of the data - it would be just as evenly spaced or clumped up, but 7 greater. The only thing that would affect the spread is multiplying every data point by 0.9. This makes distances between the data points 0.9 times as big, and thus makes the standard deviation 0.9 times as big. The standard deviation was 5.6, and so now is 5.6x0.9 = 5.04
There is no such thing as "absolute safety" or "absolute health."
No.
You can't. You need an estimate of p (p-hat) q-hat = 1 - p-hat variance = square of std dev sample size n= p-hat * q-hat/variance yes you can- it would be the confidence interval X standard deviation / margin of error then square the whole thing